Question

We are formulating a 90% confidence interval for the population mean and would like our confidence interval to have a margin of error that is not larger than 3. Experience suggests that a reasonable estimate for the population standard deviation is 15. What minimum sample size should be anticipate using? (NOTE: Round your final answer up to the next whole number).

Answer 1

A sample size of 68 should be anticipated using.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.9)/(2) = 0.05`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.05 = 0.95`, so Z = 1.645.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

Experience suggests that a reasonable estimate for the population standard deviation is 15

This means that
`\sigma = 15`

What minimum sample size should be anticipate using?

Margin of error at most 3, which means that the sample size is n when M = 3. So

`M = z(\sigma)/(√(n))`

`3 = 1.645(15)/(√(n))`

`3√(n) = 1.645*15`

Simplifying both sides by 3:

`√(n) = 1.645*5`

`(√(n))^2 = (1.645*5)^2`

`n = 67.7`

Rounding up

A sample size of 68 should be anticipated using.